Back
to top
Physics Question Helppp
Watch this thread
Announcements
Page 1 of 1
Go to first unread
Skip to page:
ChronicBoredom
Badges:
3
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
A fish floats in water with its eye at the centre of an opaque walled full tank of water of circular cross section. When the fish look upwards, it can see a fish-eye view of the surrounding scene i.e. it is able to view the hemisphere of the scene above the water surface, and centred at the top of the tank. The diameter of the tank is 30 cm, and the critical angle for water is 48 degrees. At what depth below the surface of the water, d, must the fish be floating?
Spoiler:
Show
Here's the answer, which I dont completely get. 13.5cm.
'From the fishes eye to the rim of the tank makes an angle of 48degrees with the vertical. Theemerging ray is horizontal (which is the requirement to see the horizon). Hence thedepth of the fish is, d x tan 48o = 15.0, giving d.'
I understand everything bar the 15. Where is the fifteen from??
'From the fishes eye to the rim of the tank makes an angle of 48degrees with the vertical. Theemerging ray is horizontal (which is the requirement to see the horizon). Hence thedepth of the fish is, d x tan 48o = 15.0, giving d.'
I understand everything bar the 15. Where is the fifteen from??
0
reply
Zacken
Badges:
22
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report
#2
(Original post by ChronicBoredom)
A fish floats in water with its eye at the centre of an opaque walled full tank of water of circular cross section. When the fish look upwards, it can see a fish-eye view of the surrounding scene i.e. it is able to view the hemisphere of the scene above the water surface, and centred at the top of the tank. The diameter of the tank is 30 cm, and the critical angle for water is 48 degrees. At what depth below the surface of the water, d, must the fish be floating?
A fish floats in water with its eye at the centre of an opaque walled full tank of water of circular cross section. When the fish look upwards, it can see a fish-eye view of the surrounding scene i.e. it is able to view the hemisphere of the scene above the water surface, and centred at the top of the tank. The diameter of the tank is 30 cm, and the critical angle for water is 48 degrees. At what depth below the surface of the water, d, must the fish be floating?
Spoiler:
Show
Here's the answer, which I dont completely get. 13.5cm.
'From the fishes eye to the rim of the tank makes an angle of 48degrees with the vertical. Theemerging ray is horizontal (which is the requirement to see the horizon). Hence thedepth of the fish is, d x tan 48o = 15.0, giving d.'
I understand everything bar the 15. Where is the fifteen from??
'From the fishes eye to the rim of the tank makes an angle of 48degrees with the vertical. Theemerging ray is horizontal (which is the requirement to see the horizon). Hence thedepth of the fish is, d x tan 48o = 15.0, giving d.'
I understand everything bar the 15. Where is the fifteen from??
0
reply
ChronicBoredom
Badges:
3
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
(Original post by Zacken)
Sketch the situation and identify the right-angle triangle. tan 48 = opposite/adjacent where adjancent is d and opposite is the radius = diameter/2 = 30/2 = 15.
Sketch the situation and identify the right-angle triangle. tan 48 = opposite/adjacent where adjancent is d and opposite is the radius = diameter/2 = 30/2 = 15.
I did sketch it... Why is the length of the cylinder divided by two equal to the radius of the circle. Aren't they independant things??
0
reply
ChronicBoredom
Badges:
3
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Oh wow, I didn't read the ques properly. The length of the tank is equal to the diameter of the circle (cross sectional area) of the tank I take it?
0
reply
Zacken
Badges:
22
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#5
Report
#5
(Original post by ChronicBoredom)
I did sketch it... Why is the length of the cylinder divided by two equal to the radius of the circle. Aren't they independant things??
I did sketch it... Why is the length of the cylinder divided by two equal to the radius of the circle. Aren't they independant things??
0
reply
Zacken
Badges:
22
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#6
Report
#6
(Original post by ChronicBoredom)
Oh wow, I didn't read the ques properly. The length of the tank is equal to the diameter of the circle (cross sectional area) of the tank I take it?
Oh wow, I didn't read the ques properly. The length of the tank is equal to the diameter of the circle (cross sectional area) of the tank I take it?
1
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
